First homology group of the general linear group The abelianization of the general linear group $GL(n,\mathbb{R})$, defined by $$GL(n,\mathbb{R})^{ab} := GL(n,\mathbb{R})/[GL(n,\mathbb{R}), GL(n,\mathbb{R})],$$ is isomorphic to $\mathbb{R}^{\times}$. This follows from the fact that $[GL(n,\mathbb{R}),GL(n,\mathbb{R})] \cong SL(n,\mathbb{R})$, so that $GL(n,\mathbb{R})^{ab} \cong \mathbb{R}^{\times}$ by the first isomorphism theorem.
Since for any group $G$, the homology group $H_1(G;\mathbb{Z}) \cong G^{ab}$, it turns out that $H_1(GL(n,\mathbb{R});\mathbb{Z}) \cong \mathbb{R}^{\times}$.
I wonder if it possible to obtain this result by directly computing $H_1(GL(n,\mathbb{R});\mathbb{Z})$ as the quotient $Z_1(GL(n,\mathbb{R}))/B_1(GL(n,\mathbb{R}))$ of 1-cycles by 1-boundaries.

What do $C_1$, $Z_1$ and $B_1$ look like for $GL(n,\mathbb{R})$? Is a direct computation of $H_1(GL(n,\mathbb{R}))$ possible?

In case this question turns out to be too elementary for MO, I apologize in advance. In that case I will quickly move my post to MSE. 
 A: This more of a comment, but here it goes anyway.
Following Brown's book Cohomology of Groups on page 36, we see that with respect to the standard resolution and any group $G$: $$C_2(G)\xrightarrow{\partial}C_1(G)\xrightarrow{0}\mathbb{Z}\to 0,$$ where $\partial[g|h]=[h]-[gh]+[g]$.
Consequently, $Z_1(G)=C_1(G)=\{[g]\ |\ g\in G\} $ and $B_1(G)=\{[h]-[gh]+[g]\ |\ h,g\in G\}$.
So $H_1(G)\cong G/[G,G]$.  I am not sure what more you could expect in any special case of this. 
In the case of a complex reductive (affine algebraic) group $G$, the quotient $G/[G,G]$ will always be isomorphic to an algebraic torus (isomorphic to a product of $\mathrm{GL}_1$'s), but that has nothing to do with group cohomology (it's the central isogeny theorem; see Milne's book Algebraic Groups for example).  A special case of this general fact is $G=\mathrm{GL}_n(\mathbb{C})$ where $G/[G,G]\cong \mathbb{C}^*$.  The fact that you get $\mathbb{R}^*$ for $G=\mathrm{GL}_n(\mathbb{R})$ is analogous to this case (in fact it is true for any field).
